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Momentum and Position Operator Commutator Levi Civita Form

  1. Nov 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that ##[L_i,x_j]=i\hbar \epsilon_{ijk}x_k \quad (i, j, k = 1, 2, 3)## where ##L_1=L_x##, ##L_2=L_y## and ##L_3=L_z## and ##x_1=x##, ##x_2=y## and ##x_3=z##.

    2. Relevant equations

    There aren't any given except those in the problem, however I assume we use

    $$L_i=x_jp_k-p_kx_j$$

    $$[A,B]=AB-BA$$

    $$[A+B,C]=[A,C]+[B,C]$$

    $$[AB,C]=[A,C]B+A[B,C]$$

    3. The attempt at a solution


    Sub ##L_i=x_jp_k-p_kx_j## into ##[L_i,x_j]##

    $$[x_j-p_k-p_kx_j]=[x_jp_k,x_j]-[p_kx_j,x_j]$$

    Use ##[AB,C]=[A,C]B+A[B,C]##

    $$([x_j,x_j]p_k+x_j[p_k,x_j])-([p_k,x_j]x_j+[x_j,x_j]p_k)$$

    The two commutators ##[x_j,x_j]=0## so I end up with

    $$x_j[p_k,x_j]-[p_k,x_j]x_j$$

    However I have no idea where to go from here. I don't understand where the ##x_k## term comes from. I assume the ##i\hbar## term would appear if I wrote the momentum operators as ##-i\hbar\frac{\partial}{\partial x_k}## however then I would end up with a bunch of differential operators I don't know how to work with.
     
  2. jcsd
  3. Nov 2, 2016 #2

    BvU

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    Easy: ##{\partial \over \partial x_i }x_j = \delta_{ij}##
     
  4. Nov 3, 2016 #3

    PeroK

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    This last equation is not correct. It is only true if the indices are ##(i, j, k) = (1, 2, 3), (2, 3, 1)## or ##(3, 1, 2)##. For example:
    $$L_1 = x_2 p_3 - x_3 p_2$$
    This is where the Levi-Civita symbol comes in:
    $$L_i = \epsilon_{ijk}x_j p_k$$
    Because, if ##i = 1## say, then ##\epsilon_{ijk} =0## except ##\epsilon_{123} =1## and ##\epsilon_{132} =-1##

    And note that this invokes the "summation convention", which is a convenient shorthand and hides a whole bunch of the mathematical framework. For example, written out fully that would be:
    $$For \ i = 1, 2, 3: \ \ L_i = \sum_{j,k = 1}^{3} \epsilon_{ijk}x_j p_k$$
    That is three separate equations (one for each of ##i = 1, 2, 3##) involving ##L_i## equal to a sum of 9 terms.

    My advice is that, until you have really mastered this Levi-Civita symbol, you should always keep in mind that it is an extreme shorthand and often what appears to one equation involving a couple of terms is in fact much more complicated.
     
    Last edited: Nov 3, 2016
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